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605-655 (Medium)|   Algebra|      
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Bunuel
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If l/(m + n) = m/(n + l) = n/(l + m) = k , where k is a real number 10 Sep 2021, 01:39
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72% (01:41) correct 28% (02:44) wrong based on 208 sessions

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If \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m} = k\) , where \(k\) is a real number, then which one of the following does \(k\) equal?

(A) \(\frac{1}{3}\)

(B) \(\frac{1}{2}\)

(C) 1

(D) 2

(E) 3
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B
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If l/(m + n) = m/(n + l) = n/(l + m) = k , where k is a real number 10 Sep 2021, 06:56
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Bunuel
If \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m} = k\) , where \(k\) is a real number, then which one of the following does \(k\) equal?

(A) \(\frac{1}{3}\)

(B) \(\frac{1}{2}\)

(C) 1

(D) 2

(E) 3
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We could try solving this equation, but a much faster approach is to find some values of \(l\), \(m\) and \(n\) that satisfy the given equation \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m}\)

We can see that, if \(l=1\), \(m=1\) and \(n=1\), then the equation is satisfied.

In other words, we have: \(\frac{1}{1 + 1} = \frac{1}{1 + 1} = \frac{1}{1 + 1} = k\)

So, \(k = \frac{1}{1 + 1} = \frac{1}{2}\)

Answer: B
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If l/(m + n) = m/(n + l) = n/(l + m) = k , where k is a real number 11 Oct 2024, 10:08
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Bunuel
If \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m} = k\) , where \(k\) is a real number, then which one of the following does \(k\) equal?

(A) \(\frac{1}{3}\)

(B) \(\frac{1}{2}\)

(C) 1

(D) 2

(E) 3

We could try solving this equation, but a much faster approach is to find some values of \(l\), \(m\) and \(n\) that satisfy the given equation \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m}\)

We can see that, if \(l=1\), \(m=1\) and \(n=1\), then the equation is satisfied.

In other words, we have: \(\frac{1}{1 + 1} = \frac{1}{1 + 1} = \frac{1}{1 + 1} = k\)

So, \(k = \frac{1}{1 + 1} = \frac{1}{2}\)

Answer: B
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The same doesn't work with different numbers. And the question doesn't mention about the values being same. Any other alternative solutions

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If l/(m + n) = m/(n + l) = n/(l + m) = k , where k is a real number 11 Oct 2024, 11:21
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Gautam0525
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Bunuel
If \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m} = k\) , where \(k\) is a real number, then which one of the following does \(k\) equal?

(A) \(\frac{1}{3}\)

(B) \(\frac{1}{2}\)

(C) 1

(D) 2

(E) 3

We could try solving this equation, but a much faster approach is to find some values of \(l\), \(m\) and \(n\) that satisfy the given equation \(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m}\)

We can see that, if \(l=1\), \(m=1\) and \(n=1\), then the equation is satisfied.

In other words, we have: \(\frac{1}{1 + 1} = \frac{1}{1 + 1} = \frac{1}{1 + 1} = k\)

So, \(k = \frac{1}{1 + 1} = \frac{1}{2}\)

Answer: B

The same doesn't work with different numbers. And the question doesn't mention about the values being same. Any other alternative solutions

Posted from my mobile device
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\(\frac{l}{m + n} = \frac{m}{n + l} = \frac{n}{l + m}\)

_______________________

\(\frac{l}{m + n} = \frac{m}{n + l}\) gives \(nl + l^2 =m^2 + mn\)

\(\frac{m}{n + l} = \frac{n}{l + m}\) gives \(lm + m^2 =n^2 + nl\)

\(\frac{l}{m + n} = \frac{n}{l + m}\) gives \(lm + l^2 =n^2 + mn\)

_______________________

Subtract 3 from 1:

\(nl + l^2 - lm - l^2 =m^2 + mn- n^2 - mn \)

\(nl - lm =m^2 - n^2 \)

\(l(n - m) =(m-n)(m+n)\)

\(m = n\) or \(m + m + l = 0\)

_______________________

Subtract 3 from 2:

\(lm + m^2 - lm - l^2 =n^2 + nl- n^2 - mn \)

\(m^2 - l^2 = nl - mn \)

\((m - l)(m+l) = -n(m -l)\)

\(m = l \) or \(m + n + l = 0\)

_______________________

So, \(m=n=l\) or \(m + n + l = 0\)

In the first case k = 1/2 and in the second case k = -1.
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If l/(m + n) = m/(n + l) = n/(l + m) = k , where k is a real number 13 Oct 2024, 08:48
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from the question stem :
l=(m+n)k
m=(n+l)k
n=(l+m)k

sum : l+m+n=(m+n+n+l+l+m)k=2k(l+m+n)
ie (l+m+n)(1-2k)=0
l+m+n is not = 0 because 2(l+m+n) is different than 0
so k=1/2
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If l/(m + n) = m/(n + l) = n/(l + m) = k , where k is a real number 28 Mar 2026, 11:34
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